The product of all possible values of α , for which _ x → 0 ((1 -...

Using the standard expansions for small $x$, $\cos \theta \approx 1 - \dfrac{\theta^2}{2}$ and $\sin \theta \approx \theta$.

The given limit can be written as:

$\displaystyle\lim_{x \to 0} \dfrac{1 - \left(1 - \dfrac{\alpha^2 x^2}{2}\right)\left(1 - \dfrac{(\alpha+1)^2 x^2}{2}\right)\left(1 - \dfrac{(\alpha+2)^2 x^2}{2}\right)}{((\alpha+1)x)^2}$

Neglecting higher powers of $x$, the numerator simplifies to:

$1 - \left(1 - \dfrac{x^2}{2}\left(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2\right)\right) = \dfrac{x^2}{2}\left(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2\right)$

Substituting this back into the limit:

$\displaystyle\lim_{x \to 0} \dfrac{\dfrac{x^2}{2}\left(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2\right)}{(\alpha+1)^2 x^2} = \dfrac{\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2}{2(\alpha+1)^2}$

We are given that this limit is equal to $2$. Therefore:

$\dfrac{\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2}{2(\alpha+1)^2} = 2$

$\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2 = 4(\alpha+1)^2$

Let $y = \alpha+1$. Then $\alpha = y-1$ and $\alpha+2 = y+1$. The equation becomes:

$(y-1)^2 + y^2 + (y+1)^2 = 4y^2$

$y^2 - 2y + 1 + y^2 + y^2 + 2y + 1 = 4y^2$

$3y^2 + 2 = 4y^2$

$y^2 = 2$

Substituting $y = \alpha+1$ back into the equation:

$(\alpha+1)^2 = 2$

$\alpha^2 + 2\alpha + 1 = 2$

$\alpha^2 + 2\alpha - 1 = 0$

The roots of this quadratic equation represent all possible values of $\alpha$. The product of the roots is given by $\dfrac{c}{a}$:

$\text{Product} = \dfrac{-1}{1} = -1$

Answer: $-1$